- #1

junaidnawaz

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[tex] |\gamma| \leq \cos (\beta) [/tex]

[tex] \arccos (|\gamma|) \leq \beta [/tex]

does taking the arccos() on both sides of equation changes the relational operator??

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- Thread starter junaidnawaz
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- #1

junaidnawaz

- 6

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[tex] |\gamma| \leq \cos (\beta) [/tex]

[tex] \arccos (|\gamma|) \leq \beta [/tex]

does taking the arccos() on both sides of equation changes the relational operator??

- #2

tiny-tim

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hi junaidnawaz! welcome to pf!

(have a beta: β and a gamma: γ and a ≤ )

arccos is defined as being in [0,π)

so long as β is also in [0,π), your equations are the same (because cos is monotone in that region, and therefore so is arccos)

- #3

junaidnawaz

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in my case, the range of parameters is as,

[tex] 0 \leq \beta \leq \pi /2 [/tex]

[tex]-1 \leq \gamma \leq +1 [/tex]

by taking arccos() on both-sides, would it change the operator (from [tex] \leq[/tex] to [tex] \geq[/tex] ) or would it remain same ??

- #4

tiny-tim

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by taking arccos() on both-sides, would it change the operator (from [tex] \leq[/tex] to [tex] \geq[/tex] ) or would it remain same ??

oh, i missed that!

yes, cos is decreasing, so the ≤ changes to ≥

(but, eg, sin is

- #5

junaidnawaz

- 6

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if

[tex] |\gamma| \leq \cos( \beta ) [/tex]

then

[tex] x \leq \beta[/tex]

can i find "x", by keeping the RHS fixed to [tex] \beta[/tex]

is this possible to find x ?? by keeping RHS and relation operator the same ??

- #6

tiny-tim

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sorry, i don't understand …

isn't x just |γ| ?

isn't x just |γ| ?

- #7

junaidnawaz

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When I take arccos() on both sides, it becomes

[tex] \arccos( |\gamma| ) \geq \beta [/tex]

however, i want to keep [tex] \beta [/tex] on right side, and i want to keep the relational operator as [tex] \leq [/tex], i.e.,

[tex] x \leq \beta [/tex]

what would be x ??

- #8

junaidnawaz

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I wounder if its not a stupid question .... :P

- #9

tiny-tim

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however, i want to keep [tex] \beta [/tex] on right side, and i want to keep the relational operator as [tex] \leq [/tex], i.e.,

[tex] x \leq \beta [/tex]

that's not possible (unless you replace β by some decreasing function of β, such as 1/β or -β)

- #10

junaidnawaz

- 6

- 0

Thank you :)

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